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General Relativity Fall 2019 Lecture 19: Symmetries, Spherically-Symmetric Spacetimes; Schwarzschild Solution

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General Relativity Fall 2019 Lecture 19: Symmetries, Spherically-Symmetric Spacetimes; Schwarzschild Solution General Relativity Fall 2019 Lecture 19: Symmetries, spherically-symmetric spacetimes; Schwarzschild solution Yacine Ali-Ha¨ımoud November 5, 2019 There are two regimes where GR has known analytic solutions: either in the weak-gravity regime, which we have studied so far, or in the case of highly symmetric spacetimes, on which we will now focus. The notion of symmetry is conveyed by Killing vector fields, which you have encountered in homeworks 5 and 6. µ These are vector fields K that satisfy Killing's equation K(µ;ν) = r(ν Kµ) = 0. If the Killing vector field K = @(σ∗) is the partial derivative operator with respect to some coordinate σ∗, then, in a coordinate system that has xσ∗ as σ∗ one of the coordinates, the metric components do not depend on x , i.e. @σ∗gµν = 0. FORMAL DEFINITION OF STATIONARITY, SPHERICAL SYMMETRY, HOMOGENEITY • A spacetime is stationary if it possesses a Killing vector field T(0) that is timelike in some portion of spacetime. In practice, this notion is only really useful for asymptotically flat spacetimes, in which case, the aforementioned Killing vector field should be timelike in the asymptotically flat region of spacetime. • A spacetime is spherically symmetric if it possesses 3 spacelike Killing vector fields J(1);J(2);J(3) which satisfy the commutation relations [J(i);J(j)] = −ijkJ(k): (1) j We have encountered an example of these fields in homework 6: in flat spacetime, J(i) = ijkx @(k) satisfy the commutation relations above. Expressing these in terms of spherical polar coordinates fr; θ; 'g, with the polar axis along @(3), you can easily check that J(1) = − sin '@θ − cotanθ cos '@';J(2) = cos '@θ − cotanθ sin '@';J(3) = @'; (2) Thus, the fact that J(3) is a Killing vector field translates the fact that the flat spacetime metric is invariant un- der rotations about the 3-axis. Similarly, J(1) and J(2) translate invariance under rotations about the 1 and 2 axes. Note that at any given point on the manifold, the three vectors J(1);J(2);J(3) are linearly dependent, since i i j x J(i) = ijkx x @(k) = 0. Neverthertheless, the three vector fields are linearly independent: there does not exist a linear combination with constant coefficients that vanishes everywhere. • A spacetime is homogeneous and isotropic if it is spherically symmetric, with 3 rotation generators J(i), and possesses 3 additional spacelike Killing vector fields T(1);T(2);T(3) such that [T(i);T(j)] = 0; [J(i);T(j)] = −ijkT(k): (3) You can check that, in flat spacetime, T(i) = @(i) satisfy these relations. METRIC OF A SPHERICALLY SYMMETRIC SPACETIME Using the general definition of spherical symmetry, and restricting ourselves to Killing vector fields which, at each j point of the manifold, span a 2-dimensional vector space (just like it is the case for the J(i) = ijkx @(k)), one can derive the following general form for the line element (see e.g. C. M. Hirata's lecture notes): 2 2 2 2 2 2 2 2 2 ds = gtt(t; r)dt + grr(t; r)dr + r dΩ ; dΩ ≡ dθ + sin θd' : (4) In these coordinates the three Killing vector fields J(i) take exactly the same form as in Eq. (2) { check for yourselves that these indeed satisfy the appropriate commutation relations. 2 Note that the coordinate r should not be interpreted as the \distance" to some \origin": its only clear meaning is that spacelike surfaces of constant t and r (the spheres S(P )) have an area 4πr2, assuming θ and ' span the usual range of spherical polar coordinates. It is somewhat tedious but straightforward to compute the Riemann and Ricci tensors of this metric. The non- vanishing components of the Ricci tensor are 1 2 2 1 gtt 2 2 1 Rtt = @t ln jgrrj + (@t ln jgrrj) − @t ln jgttj@t ln jgrrj − @r ln jgttj + (@r ln jgttj) − @r ln jgttj@r ln jgrrj + @r ln jgttj 2 2 grr r 1 R = @ ln jg j tr r t rr 1 2 2 1 1 grr 2 2 Rrr = − @r ln jgttj + (@r ln jgttj) − @r ln jgttj@r ln jgrrj − @r ln jgrrj − @t ln jgrrj + (@t ln jgrrj) − @t ln jgttj@t ln jgrrj 2 r 2 gtt 1 h r i Rθθ = 1 + @r ln jgrr=gttj − 1 grr 2 2 R'' = sin θRθθ BIRKHOFF'S THEOREM AND THE SCHWARZSCHILD METRIC Let us now consider the metric of a spherically-symmetric spacetime in vacuum. This does not mean that the entire spacetime must be empty: we simply focus on the vacuum regions, where Gµν hence Rµν vanishes. Setting Rtr = 0, we find that @tgrr = 0, i.e. grr(r) is a function of r only. Setting Rθθ = 0, we find that 2 @ ln jg =g j = (1 − g ); (5) r rr tt r rr which is a function of r only. Thus @r ln jgttj must be a function of r only. Integrating, we find that ln jgttj is a sum of a function of t only and a function of r only: ln jgttj = A(t) + B(r), i.e. A(t) B(r) gtt = ±e e : (6) 0 1 A(t) We can rescale t to get rid of A(t), i.e. define dt ≡ e 2 dt. This brings the metric to the following form (removing the prime on the rescaled t): 2 2 2 2 2 ds = gtt(r)dt + grr(r)dr + r dΩ : (7) We see that in this form, @t is a Killing vector field. Provided gtt < 0 in the asymptotically flat region (which we will show shortly), we have therefore already shown a non-trivial result, that a spherically-symmetric spacetime in vacuum must also be stationary! We have not yet used the tt and rr Einstein field equations. They can be combined to give grr 1 Rrr − Rtt = @r ln jgttgrrj = 0: (8) gtt r This implies that grrgtt is a constant. This constant must be negative if the metric is to have signature (−1; 1; 1; 1). We can simply rescale t by some overall multiplicative constant to make this constant −1, i.e. to have grr = −1=gtt. The last step is to finally solve for gtt. We rewrite the θθ Einstein field equation by susbtituting grr = −1=gtt: @rgtt 0 = 1 + gtt 1 + r = 1 + gtt + r@rgtt = 1 + @r(rgtt): (9) gtt Integrate this equation to get 2M rg = −r + 2M ) g = − 1 − : (10) tt tt r 3 where M is a constant of integration. We therefore arrive at the famous Schwarzschild metric 2M dr2 ds2 = − 1 − dt2 + + r2dΩ2 : (11) r 2M 1 − r To summarize, we have found that the metric of a spherically symmetric spacetime must take the Schwarzschild form in vacuum (up to coordinate redefinitions of course). This metric is only valid in vacuum, outside the sources. In particular, provided there is matter up to some r = R > 2M, we need not worry about the apparent divergence of grr at r ! 2M. We will get back to this apparent singularity later on. For r > 2M, the Killing vector field @t is timelike, hence this spacetime is stationary (that is, if spacetime was vacuum everywhere { in practice, @t need not be a Killing vector field inside the sources). For r 2M, the metric takes the following form: 2M 2M ds2 ≈ − 1 − dt2 + 1 + dr2 + r2dΩ2: (12) r r This is almost the far-field metric of an object with mass M, except that the (1+2M=r) does not multiply the angular part. To fix it, define r = r − M, so that 2M 2M 1 ± = 1 ± + O(M=r2); (13) r r 2M r2 = r2 + 2Mr + M 2 = 1 + + O(M=r2) r2: (14) r With this new coordinate, the metric at r M can therefore be rewritten as 2M 2M ds2 ≈ − 1 − dt2 + 1 + dr2 + r2dΩ2 ; (15) r r which is indeed the far-field metric of a source with mass M. TIMELIKE GEODESICS IN THE SCHWARZSCHILD METRIC Constants of motion The Schwarzschild metric has four Killing vector fields, @t and J(1);J(2);J(3). We know (homework 5) that the µ µ µ scalars K uµ are conserved along geodesics. We define E ≡ −ut = −(@t) uµ and Li ≡ J(i)uµ. Explicitly, we have dθ d' L = − sin ' u − cotanθ cos ' u = −r2 sin ' − cos θ sin θ cos ' ; (16) 1 θ ' dτ dτ dθ d' L = cos ' u − cotanθ sin ' u = r2 cos ' − cos θ sin θ sin ' ; (17) 2 θ ' dτ dτ d' L = u = g u' = r2 sin2 θ (18) 3 ' '' dτ Combining the first two equations, we find that dθ 1 = (L cos ' − L sin ') : (19) dτ r2 2 1 Suppose the orbit starts in the θ = π=2 plane, with dθ/dτ = 0, hence dθ/dτ = 0 at all times, implying that the orbit remains in the plane θ = π=2. This is true of any plane (it suffices to rotate the coordinate system to describe the plane by θ = π=2). In what follows we therefore focus on \equatorial" orbits, i.e. with θ = π=2 = constant. 2 We denote by L ≡ L3 = r dϕ/dτ. In the non-relativistic limit and for M r, we have dτ ≈ dt, and L can be interpreted as the orbital angular momentum per unit mass. To understand the meaning of E, let us write dt −1=2 E = −u = −g ut = (1 − 2M=r) = (1 − 2M=r) (1 − 2M=r) − (1 + 2M=r)(dr=dt)2 − r2(d'=dt)2 (20) t tt dτ In the limit M r and for small velocities v2 ≡ (dr=dt)2 + r2(d'=dt)2 1, we have 1 M E ≈ 1 + v2 − : (21) 2 r In other words, E can be interpreted as the energy per unit mass.
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